Confusion about getting uncertainty by using differetiation.

[kof9595995]

My question comes from my homework, but I don't think it's a homework question, so I put it here, still I will put the homework in this thread caus I think it would help.
HW problem: Show that for a free particle the uncertainty relation can also be written as
\delta \lambda \delta x \ge \frac{{{\lambda ^2}}}{{4\pi }}.
Where \delta \lambda is the de Broglie's wave length
My solution is :
\delta \lambda = |\frac{{d\lambda }}{{dp}}|\delta p = \frac{h}{{{p^2}}}\delta p
so \delta \lambda \delta x = \frac{h}{{{p^2}}}\delta x\delta p \ge \frac{h}{{{p^2}}}\frac{h}{{4\pi }} = \frac{1}{{4\pi }}{(\frac{h}{p})^2} = \frac{{{\lambda ^2}}}{{4\pi }}

Although I got the expected result, but I really doubt if it's legal to use differentiation here. Because \delta p is a standard deviation not increment. Using differentiation just means you map the range \delta p to another range \delta \lambda as if they were increments. \delta p is the standard deviation for \delta p distribution, but how do you know the \delta \lambda is the standard deviation for \delta \lambda distribution?


And I try to work out a counterexample:
Suppose W and G are two physical quantities, G follows the normal distribution
G = \frac{{10}}{{\sigma \sqrt {2\pi } }}\exp ( - \frac{{{x^2}}}{{2{\sigma ^2}}})

W = 100G = \frac{{1000}}{{\sigma \sqrt {2\pi } }}\exp ( - \frac{{{x^2}}}{{2{\sigma ^2}}})

So W and G should have the same standard deviation \sigma (I'm not quite sure , am I correct at this?), but differentiation tells you the standard deviation should be 100 times of the first distribution.

EDIT: My counterexample is wrong, please ignore it.

 

[kof9595995]

Basicly what I'm trying to ask is:
if g=g(f), and I used \delta g=\frac{{dg}}{{df}}\delta f. Then actually I presumed
\sqrt{<(g-<g>)^2>}\approx\frac{{dg}}{{df}}\sqrt{<(f-<f>)^2>}, but I can't see how to prove this.

 

[Avodyne]

You're correct that it's only approximately true. It's like approximating f(x) near x=a by f(a)+f'(a)(x-a).

 

[kof9595995]

I understand in the differentiation case why f(a)+f'(a)(x-a) is the first order approximation. But I just can't see in any way a standard deviation should behave like this, even in an approximation sense.
So \delta g=\frac{{dg}}{{df}}\delta f just seems to me more like an abuse of the notation delta here

 

[kof9595995]

Anybody can help, or is there any really correct way of doing this problem? My head really exploded, any help is appreciated.

 

[Avodyne]

Like I said, it's approximately true. It only works if g(f) varies slowly over the range in f set by the standard deviation in f. See

http://en.wikipedia.org/wiki/Propagation_of_uncertainty

 

[kof9595995]

Em, so I can use first order part of Taylor expansion to prove it, that's very helpful, thanks